3. A hanger (or a conveyor system) extends outward from two supports as shown in figure P3. The load at the right end varies from 600 lb to 3800 lb. Design the hanger.

Given: A Hanger extending outward from two supports Minimum load, Pmin = 600 lb Maximum load, Pmax = 3800 lb Required: Design the hanger Initial design decision:

Steel bar with rectangular cross-section having its long dimension vertical. Cylindrical holes drilled on the neutral axis of the bar at points A, B, and C. Cylindrical pins at holes A and B to attach the bar to the conveyor. Thickness should be fairly large to provide good bearing surface for the pins and to

ensure stability upon subjection to stress. Dimensions of the bar : Thickness, t = 1 in

Height, h = 4 in

Proposed material : AISI 1050 OQT 400 Steel Properties (from Appendix 3): Ultimate strength, SU = 143 ksi Yield strength, SY = 110 ksi

Design Factor : N = 4; Since the actual use pattern for the system is uncertain and shock loading is likely to occur

Horizontal Bar:

Analysis: >> Solving for the mean and alternating Forces, Pm and Pa:

Pm = Pmax + Pmin

2 =

3800 lb + 600 lb

2 = 2 200 lb

Pa = Pmax − Pmin

2 =

3800 lb − 600 lb

2 = 1 600 lb

>> Free-body diagram of the horizontal bar:

A B C

NEUTRAL AXIS

h

t

A B C

A B C

FA FB P

24” 36”

Solving for the reaction forces at A and B:

MB = 0 ; 24FA – 36P = 0 ; FA = 36

24P

Fy = 0 ; FA − FB + P = 0 ; FB = FA + P

Considering minimum load, Pmin = 600 lb:

FA = 36

24Pmin =

36

24 600 lb = 900 lb

FB = FA + P = 900 lb + 600 lb = 1500 lb Considering maximum load, Pmax = 3800 lb:

FA = 36

24Pmax =

36

24 3800 lb = 5700 lb

FB = FA + P = 5700 lb + 3800 lb = 9500 lb

>> Shear and Moment diagrams

Considering the minimum load

Mmin = 21 600 lb ∙ in

Considering the maximum load

Mmax = 136 800 lb ∙ in

SHEAR DIAGRAM

MOMENT DIAGRAM

- 900 lb

- 21 600 lb-in

600 lb

SHEAR DIAGRAM

MOMENT DIAGRAM

-5700 lb

- 136800 lb-in

3800 lb

A B C

FA FB Pmin

24” 36”

A B C

FA FB Pmax

24” 36”

>> Solving for the mean and alternating moment, Mm and Ma:

Mm = Mmax + Mmin

2 =

136 800 lb ∙in + 21 600 lb ∙in

2 = 79 200 lb ∙ in

Ma = Mmax − Mmin

2 =

136 800 lb ∙in − 21 600 lb ∙in

2 = 57 600 lb ∙ in

>> Solving for actual endurance strength, S n :

S n = Sn (CS)(CM )(CST )(CR)

Theoretical endurance strength, Sn

From figure 5-8, for machined steel with Su = 143 ksi, Sn ≈ 50 ksi

Size factor, CS

For the equivalent diameter, DE for rectangular cross-section,

DE = 0.808 ht = 0.808 4in 1in = 1.616 in From figure 5-9, for diameter = 1.616 in, CS ≈ 0.831

Material factor, CM

For wrought steel, CM = 1.0

Type-of-stress factor, CST

For repeated bending stress, CST = 1.0

Reliability factor, CR

For a reliability of 99.9%, CR = 0.75

So,

S n = (50 ksi) (0.831)(1.0)(1.0)(0.75) = 31.1625 ksi

>> Determining the size of the hole in the horizontal bar (size of pins)

Since the pins will experience fluctuating shear stresses (Case H), the following equation will be used for the analysis:

1

N =

τm

Ssu +

τa

S sn

Where:

τm =Pm

2A ; τm =

Pa

2A ; Ssu = 0.75Su ; S sn = 0.577S n

1

N =

τm

Ssu +

τa

S sn =

P m2A

0.75Su +

P a2A

0.577S n =

1

2A

Pm

0.75Su +

Pa

0.577S n

Solving for the cross-sectional area of the pin:

A = N

2

Pm

0.75Su +

Pa

0.577S n =

4

2

2 200 lb

0.75 143000 lb

in 2 +

1600 lb

0.577 31162 .50 lb

in 2

A = 0.2190 in2 Solving for the minimum allowable diameter of the pin:

A = 0.2190 in2 = π

4D2

D = 4A

π =

4 0.2190 in 2

π = 0.5281 in

Since the diameter is quite small considering the loads and other stress like the bearing stress and wear at the surfaces that contact the tabs of the bar, a larger diameter would be preferred. Specification: Use D = 0.75 in

>> Stress concentration factor, kt

From figure A15 – 3, for flat plate with a central hole,

Assuming d

h ≤ 0.50, Use kt = 1.0

>> Determining the section modulus of the cross-sectional area of the bar Solving for the minimum section modulus, Smin:

Since the bar will experience fluctuating normal stress due to bending (Case G), the Goodman Method is applicable. The following equation will be used for the analysis:

1

N =

σm

Su +

ktσa

S n

Where:

σm =Mm

Smin and σm =

Ma

Smin

1

N =

M mSmin

Su +

k t M aSmin

S n =

1

Smin

Mm

Su +

kt Ma

S n

Smin = 1

N

Mm

Su +

kt Ma

S n =

1

4

79 200 lb ∙in

143000 lb

in 2

+ 1.0 57 600 lb ∙in

31162 .50 lb

in 2

Smin = 0.6006 in3

Solving for the required height section, h:

For solid rectangular section, S = th3

6

h = 6S

t =

6 0.6006 in 3

1.0 in = 1.8983 in ≈ 1.90 in

The table of preferred basic sizes in the decimal-inch system, Table A2 – 1, recommends h = 2.0 in.

Since the actual height is not identical to the assumed height, h = 4 in, the solution is not acceptable.

Recalculating the height with all the related parameters:

DE = 0.808 ht = 0.808 2in 1in = 1.14 in ; CS ≈ 0.865

S n = (50 ksi) (0.865)(1.0)(1.0)(0.75) = 32.4375 ksi

Smin = 1

4

79 200 lb ∙in

143000 lb

in 2

+ 1.0 57 600 lb ∙in

32437 .50 lb

in 2

= 0.5824 in3

h = 6S

t =

6 0.5824 in 3

1.0 in = 1.8693 in ≈ 1.87 in

Hence, for the actual height, use h = 2.0 in

Recalculating the pin diameter:

A = 4

2

2 200 lb

0.75 143000 lb

in 2 +

1600 lb

0.577 32437 .50 lb

in 2 = 0.2120 in2

D = 4A

π =

4 0.2120 in 2

π = 0.5195 in

Use D = 0.75 in

Solving for the actual ratio of the diameter and height at point C, to check if the

earlier assumption d

h < 0.50 is correct,

d

h =

0.75 in

2 in = 0.375

Since the actual value satisfies the assumed value, then the earlier

assumption kt = 1.0 , is acceptable.

Solving for the actual section modulus of the cross section with the holes in it,

S = t h3 − d3

6h =

1 in 2 in 3 – 0.75 in 3

6 2 in = 0.6315 in3 ≈ 0.63 in3

Since the actual value is larger than the minimum required value, the size of the cross section is satisfactory with regard to stress due to bending.

>> Therefore,

The following are the final design decisions for the horizontal bar of the hanger (or conveyor system): Material : AISI 1050 OQT 400 Steel Size : Rectangular cross-section

Section Modulus, S = 0.63 in3

Thickness, t = 1 in Height, h = 2 in Hole diameter = 0.75 in (for the pins)